Viewing a single comment thread. View all comments

The_Illist_Physicist t1_jagtuez wrote

Seeing as though nobody is really answering your question I'll take a swing at things. As a disclaimer, while I am a student of physics, my specialty is optics and not fluid dynamics.

From the Wikipedia page on drag coefficients, it looks like the total drag coefficient (C_d) is defined as the sum of pressure drag (C_p) and friction drag (C_f). These can be thought of respectively as:

The component of drag that comes from the traditional resistive force we think of due to the fluid exerting a pressure on the face of the projectile, and a "negative pressure" behind the projectile. If you look at the expression to compute this, there's a dot product between the free stream flow unit vector and the negative surface element normal unit vector (inside a surface integral). What this means is for a cube in a non-turbulent fluid, only the front and back faces of the cube contribute to this C_p term. By thinking about the symmetry, a spinning bullet should have no effect here.

The component of drag that comes from a shear stress due to fluid friction along the sides of the projectile. Again in this surface integral there's a dot product, this time between the shear stress unit vector and and the free stream flow unit vector. This is then scaled by multiplying it by the total shear stress at the differential surface element. For a spinning projectile, this total shear stress value is larger than if it was not spinning, so by this reasoning a spinning bullet should experience more drag than a stable, non rotating bullet.

Now I suppose this is all for a very idealized scenario. However a quick search for literature on friction drag simulations turned up this paper https://www.hindawi.com/journals/ijae/2020/6043721/ which seems to suggest that the spinning effect does indeed have some small contribution to the total drag, although it gets becomes less important at higher speeds (below Mach 1).

TL;DR: Yes it appears a bullet does incur a little bit of extra drag from it spinning, however it's necessary to stabilize the bullet and is fairly negligible when the projectile speed is high.

400

MassProductionRagnar t1_jahf5wj wrote

>and a "negative pressure" behind the projectile.

And quite a lot. For an artillery shell this can be as high as 40% of the drag. Modern artillery shells have pyrotechnic charges at the back to reduce this. They aren't pushing the shell, they just fill up the empty void with gases, reducing the drag by a lot. That type of ammunition called base-bleed can have up to 30% longer range.

E.g. a German PzH2000 self-propelled howitzer firing a base-bleed shell in Ukraine:

https://pbs.twimg.com/media/FoVlKRDWAAACjsY?format=jpg&name=large

80

The_Illist_Physicist t1_jahksa4 wrote

Very interesting, I wasn't aware this effect was so substantial! Our capacity for making weapons to kill each other is truly remarkable.

39

MyNameIsRay t1_jaig3gl wrote

A lot of long-range shooting enthusiasts use projectiles known as "boat tails" for the same reason.

Instead of having a flat back, they taper down a bit.

By making the surface area of the rear smaller, the size of the negative pressure zone is also made smaller, reducing the effect and raising the ballistic coefficient.

11

scheav t1_jahunkm wrote

“Filling the empty void with gases” IS pushing the shell. It might not have a net positive acceleration, but it’s like pushing a car that’s out of gas - you’re still pushing even though it’s not accelerating.

−5

ramk13 t1_jahw8on wrote

I think their point was that the purpose wasn't propulsion, but a reduction in drag. If you look at the momentum imparted by the gas released it will be much smaller than the decrease in momentum caused by drag if the gas weren't there.

29

MassProductionRagnar t1_jahwc4l wrote

>IS pushing the shell

By a negliglible amount. There are actual range-boosted shells with rocket motors which gain significant thrust from the shells actually accelerated in flight.

Base-bleed is not that and for simplicities sake these shells aren't said to be propelled by it, they just have less drag.

18

haplo_and_dogs t1_jai6e2x wrote

Vacuum doesn't exert a force. The gases released do.

You can look at this from a force diagram. Having a vacuum behind the shell just means there isn't gas pushing on the shell. Adding a gas does push on the shell.

5

Phoenixdive t1_jaje6is wrote

Out of curiosity, in your force diagram, what happens to the matter around a vacuum? What forces are created by the pressure differential? If you found a way to decrease the pressure differential, what would happen to those forces? Which one would exert a greater net gain in speed, the equalization of the pressure differential or the ejection of mass to fill said void in this particular scenario?

1

haplo_and_dogs t1_jajjp4n wrote

>what happens to the matter around a vacuum?

Matter in a vacuum has no forces exerted on it.

>What forces are created by the pressure differential?

Pressure is just Force / Area.

Pressure ( in this example ) is the result of the summation of forces of gases bouncing off of a surface. There is very high pressure at the front of the shell, and very low pressure at the rear of a shell when fired.

>If you found a way to decrease the pressure differential, what would happen to those forces?

You can reduce drag if you change the shape of the shell. If you have more pressure at the rear of the shell the net force on the shell is reduced. You can do this with a boat-tail shell.

>Which one would exert a greater net gain in speed, the equalization of the pressure differential or the ejection of mass to fill said void in this particular scenario?

The gas sent out of the rear of the shell does not act like a rocket. In a bleed gas shell the gas is going nearly the same speed as the shell, and is trapped there for multiple interactions. The trapping of this allows the gas to exert a pressure onto the shell. If your force diagram model has this gas as part of the shell you have a new drag coeffienct, if you model the shell alone then you have this gas exerting a force ( or a pressure over an area ) on to the shell.

2

scheav t1_jaiaxfq wrote

I didn’t say they are propelled by it, I said they are pushed by it. It exerts a force, which reduces its acceleration due to drag. This is called pushing.

2

interstellargator t1_jah5vew wrote

Just to clarify, is that additional drag in the direction of travel? Or is it the same amount in the direction of travel plus a small lateral amount in the direction of spin?

13

MichelanJell-O t1_jah78bk wrote

Excellent point. The latter. If you divide the motion of the bullet into a linear component and a rotational component, the additional friction drag only acts against the rotation. So the spinning bullet will slow its rotation over time, but the two bullets should have almost exactly the same speed.

12

elPocket t1_jah8m62 wrote

There is another effect that can result in additional spin drag in the form of viscous drag.

Preamble: both bullets fly with the exact same speed and perfect orientation, so zero angle of attack.

As you elaborated, the viscous friction in the spinning case is higher. This can be additionally contributed by an earlier boundary transition from laminar to turbulent.

In the spinning case, the flow spirals around the bullet, effectively traveling a longer distance in the same amount of time. The relative velocity between surface and air is higher. This increases the Reynolds number, (rho * v * l_ref / eta; rho, l_ref & eta are constant between the cases) possibly leading to earlier laminar-turbulent transition. Also, the flow distance along the spiral is bigger, so there's more distance for the transition to happen. If it were to happen at the same run length of the boundary layer, due to the spiral the transition would effectively happen earlier along the bullet spin axis. And since turbulent boundary layers exert more drag than laminar ones, this can increase drag for the spinning bullet.

12

uiucengineer t1_jaib6kq wrote

>For a spinning projectile, this total shear stress value is larger than if it was not spinning, so by this reasoning a spinning bullet should experience more drag than a stable, non rotating bullet.

But the direction of this increase in drag should be orthogonal to the translational motion, no? If I'm right, this drag would contribute to slowing the rotation, but not to the arc of the projectile.

1